Braking distance

The car travels at an average speed of 12 km/h and detects an obstacle 10 m in front of it. At 1 m in front of the obstacle it already runs 2 km/h. What is the braking distance? What is the required deceleration for stop in:

A) 1m
B) 1s?

Correct result:

x =  12.96 m
a₂ =  0.5556 m/s²

Solution:

$$\begin{gathered} v_1=12\text{ km/h}\to \text{ m/s}=12\mathrm{/}3.6\text{ m/s}=3.33333\text{ m/s } \\ {s}_1=10\text{ m } \\ {s}_2=1\text{ m } \\ {v}_2=2\text{ km/h}\to \text{ m/s}=2\mathrm{/}3.6\text{ m/s}=0.55556\text{ m/s } \\ s=s_1-s_2=10-1=9\text{ m } \\ s={\displaystyle \frac{1}{2}}a{t}^2 \\ at=(v_1-v_2) \\ s={\displaystyle \frac{1}{2}}(v_1-v_2)\cdot t \\ t=2\cdot s\mathrm{/}(v_1-v_2)=2\cdot 9\mathrm{/}(3.3333-0.5556)={\displaystyle \frac{162}{25}}=6.48\text{ s } \\ a={\displaystyle \frac{v_1-v_2}{t}}={\displaystyle \frac{3.3333-0.5556}{6.48}}={\displaystyle \frac{625}{1458}}\doteq 0.4287{\text{m/s}}^2 \\ {t}_1=v_1\mathrm{/}a=3.3333\mathrm{/}0.4287={\displaystyle \frac{972}{125}}=7.776\text{ s } \\ x={\displaystyle \frac{1}{2}}\cdot a\cdot t_1^2={\displaystyle \frac{1}{2}}\cdot 0.4287\cdot 7.776^2={\displaystyle \frac{324}{25}}={\displaystyle \frac{324}{25}}\text{ m}=12.96\text{ m} \end{gathered}$$

$$\begin{gathered} t_2=2\cdot s_1\mathrm{/}(v_1-0)=2\cdot 10\mathrm{/}(3.3333-0)=6\text{ s } \\ {a}_2={\displaystyle \frac{v_1}{t_2}}={\displaystyle \frac{3.3333}{6}}={\displaystyle \frac{5}{9}}=0.5556{\text{m/s}}^2 \end{gathered}$$

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